\documentclass[11pt]{scrartcl}
\usepackage{evan}

\begin{document}
\title{H3226403}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 138}
\maketitle

\section*{Problem}
Let $ABC$ be a triangle with circumcircle $\Omega$ and orthocenter $H$.
Let the midpoint of $AH$ be $N$ and let the reflections of $N$
over $\ol{AC}$, $\ol{AB}$ be $N_1$, $N_2$, respectively.
Let the line perpendicular to $N_1N_2$ through $N_1$ intersect $AC$ at $K$
and let the line perpendicular to $N_1N_2$ through $N_2$ intersect $AB$ at $L$.
The circumcircle of triangle $AKL$ intersect $\Omega$ at $A$ and $X$;
let $HX$ intersect $\Omega$ at $X$ and $Y$.
Show that $\ol{BY} \parallel \ol{HL}$ and $\ol{CY} \parallel \ol{HK}$.

\section*{Video}
\href{https://www.youtube.com/watch?v=YTRMS2MG8P0&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/YTRMS2MG8P0}}

\section*{External Link}
\url{https://aops.com/community/p29539773}

\newpage

\section*{Solution}
Let $\ol{BH}$ and $\ol{CH}$ meet $\Omega$ again at $B'$ and $C'$, respectively.
It's known that $\ol{AO} \perp \ol{B'C'}$,
and these two lines meet at the midpoint $M$ of $\ol{B'C'}$.
Ignore points $X$ and $Y$ for now.

\begin{center}
\begin{asy}
/*
Converted from GeoGebra by User:Azjps using Evan's magic cleaner
https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py
*/
pair A = (-3.02464,5.89564);
pair B = (-5.,-0.5);
pair C = (1.5,-0.5);
pair O = (-1.75,1.99908);
pair H = (-3.02464,0.89748);
pair N_1 = (-0.66809,5.06371);
pair N_2 = (-4.43393,3.83183);
pair K = (0.85251,0.41522);
pair L = (-4.03692,2.61817);
pair X = (2.34961,2.03161);
pair Y = (-5.51318,0.37232);
pair Bp = (1.68844,4.23179);
pair Cp = (-5.84322,1.76802);
pair M = extension(A,O,Bp,Cp);

import graph;
size(12cm);
pen zzttqq = rgb(0.6,0.2,0.);
pen qqffff = rgb(0.,1.,1.);
pen qqwuqq = rgb(0.,0.39215,0.);
pen yqqqqq = rgb(0.50196,0.,0.);
pen xdxdff = rgb(0.49019,0.49019,1.);
pen cqcqcq = rgb(0.75294,0.75294,0.75294);
draw(A--B--C--cycle, linewidth(0.6) + zzttqq);

draw(A--B, linewidth(0.6) + zzttqq);
draw(B--C, linewidth(0.6) + zzttqq);
draw(C--A, linewidth(0.6) + zzttqq);
draw(K--N_1, linewidth(0.6) + qqffff);
draw(L--N_2, linewidth(0.6) + qqffff);
draw(N_1--N_2, linewidth(0.6) + qqffff);
draw(circle((-0.74522,3.39656), 3.38247), linewidth(0.6) + qqwuqq);
draw(circle(O, 4.09974), linewidth(0.6) + yqqqqq);
draw(H--L, linewidth(0.6) + qqwuqq);
draw(H--K, linewidth(0.6) + qqwuqq);
draw(K--L, linewidth(0.6) + qqwuqq);
draw(C--Cp, linewidth(0.6) + yqqqqq);
draw(B--Bp, linewidth(0.6) + yqqqqq);
draw(A--Bp, linewidth(0.6) + xdxdff);
draw(A--Cp, linewidth(0.6) + xdxdff);
draw(A--O, linewidth(0.6) + qqffff);
draw(A--(-3.02464,-1.89748), linewidth(0.6) + yqqqqq);
draw(Bp--Cp, linewidth(0.6));
draw(X--Y, dotted+grey);

dot("$A$", A, dir((-10.935, 16.900)));
dot("$B$", B, dir((-24.153, -28.485)));
dot("$C$", C, dir((12.784, -34.450)));
dot("$O$", O, dir((3.760, 7.913)));
dot("$H$", H, dir((3.977, 7.726)));
dot("$N_1$", N_1, dir((3.928, 7.639)));
dot("$N_2$", N_2, dir((-36.024, 12.527)));
dot("$K$", K, dir((35.780, -19.602)));
dot("$L$", L, dir((-33.972, -13.237)));
dot("$X$", X, dir(0));
dot("$Y$", Y, dir((-34.470, -18.294)));
dot("$B'$", Bp, dir((3.881, 8.319)));
dot("$C'$", Cp, dir((-37.255, -3.775)));
dot("$M$", M, dir(70));
\end{asy}
\end{center}

\begin{claim*}
  Point $L$ is the circumcenter of $\triangle MHC'$.
  Similarly, point $K$ is the circumcenter of $\triangle MHB'$.
\end{claim*}
\begin{proof}
  The perpendicular bisector of $\ol{HC'}$ is line $MB$
  while the perpendicular bisector of $\ol{C'M}$ is the line through $N_2$
  perpendicular to $\ol{C'M} \parallel \ol{N_1N_2}$.
  So these two perpendicular bisectors indeed meet at $L$.
\end{proof}

\begin{claim*}
  Points $K$, $H$, $L$, $A$ are cyclic.
\end{claim*}
\begin{proof}
  We see $\ol{KL}$ is the perpendicular bisector of $\ol{MH}$.
  Then, it follows that
  \[ \dang MHL = 90\dg - \dang HC'M = 90\dg - \dang CC'B'
    = 90 \dg - \dang CBB' = \dang ACB. \]
  Similarly, $\dang KHM = \dang CBA$.
  So $\dang KHL = \dang KHM + \dang MHL = \dang CAB = \dang KAL$.
\end{proof}

\begin{remark*}
  Alternatively, trig Ptolemy provides a straightforward way to prove
  the second claim without having noticed the circumcenters from before.
\end{remark*}

\begin{remark*}
  In fact, quadrilateral $ALHK$ is harmonic too.
  Indeed, $\ol{AO}$ is the median of $\triangle ALK$,
  because line $\ol{AO}$ is halfway between parallel lines $\ol{LN_2}$ and $\ol{N_1K}$,
  And lines $\ol{AO}$ and $\ol{AH}$ are isogonal in $\angle LAK$.

  So actually $M$ is the $A$-Humpty point of $\triangle AKL$,
  while $N$ is the $A$-Dumpty point.
  However, we will not use this in what follows.
\end{remark*}

We finally turn our attention to $X$ and $Y$ which we previously ignored.
To finish, we simply use Reim's theorem in the form
\[ \dang BYH = \dang BYX = \dang BAX = \dang LAX = \dang LHA. \]

\end{document}
